Problem on MM equation
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
Expert
According to M.M Equation, V = Vmax x {S]/Km + {S] Since Vmax = ksec^-1 {Et}V = Ksec^-1{Et} x {S}/Km + {S}Substituting the given value of Ksec^-1 in the above equation,we get V = Vmax x mol.wt x 1min/60 sec x {S}/mg of enzyme / Km +{S]V = Vmax x mol.wt x {S}/mg of enzyme/Km +{S}Substituting the given value of {S}V = Vmax x mol.wt x 4.576(log K -10.753-logT+Ea/4.576T)/mg of enzyme/ Km + 4.576(log K -10.753-logT+Ea/4.576T)
Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.
I want it before 8 am tomorow please. I am just wondering how much is going to be ?
Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately
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